Question

find the equation to the right circular cylinder of radius 2 whose axis passes through (1,2,3), and
has direction cosines proportional to 2, -3,6.​

Answer 1

Step-by-step explanation:

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Answer 2

Concept Introduction:-

It could take the shape of a word or a numerical representation of a quantity's arithmetic value.

Given Information:-

We have been given that radius $2$ whose axis passes through $(1,2,3)$, and

has direction cosines proportional to $2, -3,6$.​

To Find:-

We have to find that the equation to the right circular cylinder

Solution:-

According to the problem

The equation of the axis of the cylinder

$\frac{x-1}{2}= \frac{y-2}{-3}=\frac{z-3}{6}....(i)\\Distance=\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}\\ Radius=2\\Equation of line=\frac{2(x-1)+(-3)(y-2)+(6)(z-3)}{\sqrt{2^2+(-3)^2+6^2}}\\ =\frac{2x-3y+6z-14}{\sqrt{4+9+36}}\\ =\frac{2x-3y+6z-14}{\sqrt{49}}\\ =\frac{2x-3y+6z-14}{7}$

According to pyhthagoras theorem,

$(\sqrt{(x-1)(y-2)(z-3)})^2\\=(\frac{(2x-3y)(6z-14)}{\sqrt{49}})^2+2^2\\=(x^2-2x+1)+(y^2-4y+4)+(z^2-6z+9)\\=\frac{(2x-3y)^2+(6z-14)^2+2(2x-3y)(6z-14)}{49}+4\\ =x^2+y^2+z^2-2x-4y-6z+14-4\\=\frac{(4x^2+9y^2-12xy)+(36z^2+196-168z)+2(12zx-28x-18yz+42y)}{49} \\=49(x^2+y^2+z^2-2x-4y-6z+10)=4x^2+9y^2+36z^2-12xy-168z+24zx-56x-36yz+84y+196\\=49x^2+49y^2+49z^2-98x-196y-294z+490=4x^2+9y^2+36z^2-12xy-168z+24zx-56x-36yz+84y+196\\=45x^2+40y^2+13z^2+36yz-24zx+12xy-42x-280y-126z+294=0$

Final Answer:-

The equation to the right circular cylinder is $45x^2+40y^2+13z^2+36yz-24zx+12xy-42x-280y-126z+294=0$.

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